Math Problem Statement
Solve the electric field strength of the capacitor's dielectric when the polarization P=2μCm2P=2μCm2, the relative permittivity of the dielectric is εr=5.5εr=5.5 and when the permittivity of the vacuum is known ε0=8.854∗10−12Fm
Solution
To solve for the electric field strength in the dielectric of a capacitor, given the polarization , relative permittivity , and permittivity of vacuum , we can use the following relation:
Where:
- is the polarization,
- is the vacuum permittivity,
- is the relative permittivity of the dielectric, and
- is the electric field strength.
We want to solve for :
Now, let's plug in the values:
Now let's calculate .The electric field strength in the dielectric is approximately .
Would you like more details or have any questions about this solution?
Here are five related questions for further exploration:
- How does the electric field strength change with varying polarization ?
- What effect does increasing the relative permittivity have on the electric field strength?
- How does the electric field strength compare in a vacuum versus in a dielectric?
- Can you calculate the electric field strength if the relative permittivity is doubled?
- What is the relationship between capacitance and the electric field in a dielectric material?
Tip: The electric field inside a dielectric material is always weaker than in a vacuum, due to the dielectric's ability to polarize and reduce the field strength.
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Math Problem Analysis
Mathematical Concepts
Electromagnetism
Electric Fields
Capacitors
Dielectrics
Formulas
P = ε₀(εr - 1)E
E = P / [ε₀(εr - 1)]
Theorems
Gauss's Law for Electric Fields in Dielectrics
Suitable Grade Level
College-Level Physics
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